Problem: $f(x) = \dfrac{ 5 }{ \sqrt{ 4 - \lvert x \rvert } }$ What is the domain of the real-valued function $f(x)$ ?
Solution: First, we need to consider that $f(x)$ is undefined anywhere where the radicand (the expression under the radical) is less than zero. So we know that $4 - \lvert x \rvert \geq 0$ This means $\lvert x \rvert \leq 4$ , which means $-4 \leq x \leq 4$ Next, we need to consider that $f(x)$ is also undefined anywhere where the denominator is zero. So we know that $\sqrt{ 4 - \lvert x \rvert } \neq 0$ , so $\lvert x \rvert \neq 4$ This means that $x \neq 4$ and $x \neq -4$ So we have four restrictions: $x \geq -4$ $x \leq 4$ $x \neq -4$ , and $x \neq 4$ Combining these four, we know that $x > -4$ and $x < 4$ ; alternatively, that $-4 < x < 4$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid -4< x <4\, \}$.